Tim Wescott wrote:

snip

What this means is that in general your transmitter, like your wall socket,
can deliver more than it's rated power if you put the right load on it.
Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it. Your
circuit doesn't act exactly like that, even if it doesn't follow the same
curve for a generator with a linear impedance. I suspect that with your
circuit the 7-pole filter is evening things out quite a bit. There's also a
good chance that the "strange results" you see with a load below 10 ohms are
the final amplifier going unstable and oscillating, or your output
transistor heating up and changing characteristics on you.

Yes there were bad things going on with very low value loads, the output was
collapsing and re-establishing. The srtange results may also be caused by the
loads with were high wattage resistors - no idea what inductance they might
have had.

So what it boils down to is that it is very important that your output stage
_sees_ the right impedance, but you shouldn't expect the output stage to
_deliver_ the same impedance that it needs to see. As long as you're
getting power out and your output transistor isn't letting all the smoke out
then you're fine.

So perhaps the fact that the transmitter is connected to a 50 ohm load and the
output transformer is 2:1 turns ratio means that the transistor is "seeing"
200 ohms?

Nevertheless I am still interested in how to verify that any circuit which I
design (copy!) does have a 50 ohm output impedance. There must be some way to
verify this figure.

--Gary

Richard Hosking wrote:

As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po
is output power and Vcc is the supply voltage
It looks like your measurements point to 50 ohms as about right. How were
you measuring the voltage? - this might affect the measurement


Vcc * Vcc / Po = 13.8 * 13.8 / .352 = 540 ohms (for a 47 ohm load).

This value is certainly in the right ballpark.

I guess that I should also measure the current taken by the final stage and
see if this ties in too.

I connected various 1/4w resistors across the output using a couple of
adaptors, but being aware to keep wire lengths as short as possible. I then
simply used a scope probe and scope to make the measurement. Scope probe was a
standard x10 10Mohm impedance and presumably small C.

Interestingly I was surprised at the magnitude of the voltages for such a tiny
power.

regards..

--Gary

Richard Hosking wrote:

As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po
is output power and Vcc is the supply voltage
It looks like your measurements point to 50 ohms as about right. How were
you measuring the voltage? - this might affect the measurement


Vcc * Vcc / Po = 13.8 * 13.8 / .352 = 540 ohms (for a 47 ohm load).

This value is certainly in the right ballpark.

I guess that I should also measure the current taken by the final stage and
see if this ties in too.

I connected various 1/4w resistors across the output using a couple of
adaptors, but being aware to keep wire lengths as short as possible. I then
simply used a scope probe and scope to make the measurement. Scope probe was a
standard x10 10Mohm impedance and presumably small C.

Interestingly I was surprised at the magnitude of the voltages for such a tiny
power.

regards..

--Gary

Paul Burridge wrote:
On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:


Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.

Leon
--
Leon Heller, G1HSM
Email:
My low-cost Philips LPC210x ARM development system:
http://www.geocities.com/leon_heller/lpc2104.html

Paul Burridge wrote:
On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:


Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.

Leon
--
Leon Heller, G1HSM
Email:
My low-cost Philips LPC210x ARM development system:
http://www.geocities.com/leon_heller/lpc2104.html

On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:



Paul Burridge wrote:
On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:


Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.

Hi, Leon,

Do you know Brian (G4BCO) by any chance?
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.
--

"I expect history will be kind to me, since I intend to write it."
- Winston Churchill

On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:



Paul Burridge wrote:
On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:


Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.

Hi, Leon,

Do you know Brian (G4BCO) by any chance?
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.
--

"I expect history will be kind to me, since I intend to write it."
- Winston Churchill

Paul Burridge wrote:

On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:



Paul Burridge wrote:

On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:



Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.


Hi, Leon,

Do you know Brian (G4BCO) by any chance?
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.

No, that isn't the case. From AN721:

"The load value is primarily dictated by the required output power and
the peak voltage; it is not matched to the output impedance of
the device."

Leon
--
Leon Heller, G1HSM
Email:
My low-cost Philips LPC210x ARM development system:
http://www.geocities.com/leon_heller/lpc2104.html

Paul Burridge wrote:

On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:



Paul Burridge wrote:

On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:



Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.


Hi, Leon,

Do you know Brian (G4BCO) by any chance?
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.

No, that isn't the case. From AN721:

"The load value is primarily dictated by the required output power and
the peak voltage; it is not matched to the output impedance of
the device."

Leon
--
Leon Heller, G1HSM
Email:
My low-cost Philips LPC210x ARM development system:
http://www.geocities.com/leon_heller/lpc2104.html

OK, my point is (and you can test this yourself), that (1) the concept of Rs
is invalid for a class C stage, because it's not a linear device. You can
fool yourself into thinking it is, but it's not. (2) the typical RF output
stage, sans output filter, which consists of a transistor fed by a choke or
transformer and which pulls the collector voltage nearly down to the emitter
acts like an RF voltage source with a fairly low series resistance. I've
_done_ this. I've _measured_ this. I've _burnt_ transistors.

You _can_ get much more power from your output stage than it's "normal"
output power, for that brief (and often shining) moment before the output
transistor lets the smoke out and reverts to the sand and basic
petrochemicals from which it was originally made. This is how you know that
you aren't running it at the Rs = Rl point -- rather, you're running it at
the point where the junction temperature of the transistor is kept to a
level that will let it stay a transistor for a satisfying amount of time.

This is why in my original reply I compared the output amp to a wall socket.
The equivalent Rs of your average wall socket is probably on the order 1/2
to 4 ohms (on the low end for 120V, high for 230). Since I live in the US
and have fairly good feed, I'll use the 1-ohm 120V case, and I'll match Rs
with Rl. Lessee. 120V, 1-ohm, that's 14400 watts -- good! But my circuit
is only is only good for 15 amps and I'm asking for 120?!? Dang! What
happened to the lights?

So you see, every time you plug anything into the wall, you're failing to
match Rs with Rl, and that's a _very_ good thing. It's for exactly the same
reason that you don't match Rs with Rl in your output stage.

"Paul Burridge" wrote in message
...
On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:
-- snip --
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.
--

"I expect history will be kind to me, since I intend to write it."
-
Winston Churchill